Question: Simplify the following expression: $\dfrac{88r^4}{55r^2}$ You can assume $r \neq 0$.
Answer: $ \dfrac{88r^4}{55r^2} = \dfrac{88}{55} \cdot \dfrac{r^4}{r^2} $ To simplify $\frac{88}{55}$ , find the greatest common factor (GCD) of $88$ and $55$ $88 = 2 \cdot 2 \cdot 2 \cdot 11$ $55 = 5 \cdot 11$ $ \mbox{GCD}(88, 55) = 11 $ $ \dfrac{88}{55} \cdot \dfrac{r^4}{r^2} = \dfrac{11 \cdot 8}{11 \cdot 5} \cdot \dfrac{r^4}{r^2} $ $\phantom{ \dfrac{88}{55} \cdot \dfrac{4}{2}} = \dfrac{8}{5} \cdot \dfrac{r^4}{r^2} $ $ \dfrac{r^4}{r^2} = \dfrac{r \cdot r \cdot r \cdot r}{r \cdot r} = r^2 $ $ \dfrac{8}{5} \cdot r^2 = \dfrac{8r^2}{5} $